For $A \neq 0, x

Question

For $A \neq 0, x < 0, \lim _{n \rightarrow \infty} \frac{\sin x-e^{n x}}{1+A e^{n x}}=$

Aicharya · Accepted Answer

Given, $\lim _{n \rightarrow \infty} \frac{\sin x-e^{n x}}{1+A e^{n x}}=\lim _{n \rightarrow \infty} \frac{e^{n x}\left(\frac{\sin x}{e^{n x}}-1\right)}{e^{n x}\left(\frac{1}{e^{n x}}+A\right)}$ $=\lim _{n \rightarrow \infty} \frac{\frac{\sin x}{e^{n x}}-1}{\frac{1}{e^{n x}}+A}=\frac{0-1}{0+A}=-\frac{1}{A}$

For \(A \neq 0, x < 0, \lim _{n \rightarrow \infty} \frac{\sin x-e^{n x}}{1+A e^{n x}}=\)

Solution